Answer:
163.2g
Explanation:
First let us generate a balanced equation for the reaction. This is shown below:
4Al + 3O2 —> 2Al2O3
From the question given, were were told that 3.2moles of aluminium was exposed to 2.7moles of oxygen. Judging by this, oxygen is excess.
From the equation,
4moles of Al produced 2moles of Al2O3.
Therefore, 3.2moles of Al will produce = (3.2x2)/4 = 1.6mol of Al2O3.
Now, let us covert 1.6mol of Al2O3 to obtain the theoretical yield. This is illustrated below:
Mole of Al2O3 = 1.6mole
Molar Mass of Al2O3 = (27x2) + (16x3) = 54 + 48 =102g/mol
Mass of Al2O3 =?
Number of mole = Mass /Molar Mass
Mass = number of mole x molar Mass
Mass of Al2O3 = 1.6 x 102 = 163.2g
Therefore the theoretical of Al2O3 is 163.2g
Answer:
the anserw should be 665KJ
It is called DISPLACEMENT of the object....
Answer:
7.5 moles of CaBr2 are produced
Explanation:
Based on the equation:
2AlBr3 + 3CaO → Al2O3 + 3CaBr2
<em>2 moles of AlBr3 produce 3 moles of CaBr2 if CaO is in excess.</em>
<em />
Using this ratio: 2 moles AlBr3 / 3 moles CaBr2. 5 moles of AlBr3 produce:
5 moles AlBr3 * (3 moles CaBr2 / 2 moles AlBr3) =
<h3>7.5 moles of CaBr2 are produced</h3>
<em />
